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Six SHSAT Questions

3 English and 3 math

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Question 1 of 6

2026 Practice Test C English, question 24
Passage Bird Talk

Bird Talk

by Carl Sandburg

And now when the branches were beginning to be heavy,
It was the time when they once had said, “This is the
  beginning of summer.”
The shrilling of the frogs was not so shrill as in the
  first weeks after the broken winter;
The birds took their hops and zigzags a little more
  anxious; a home is a home; worms are worms.
The yellow spreads of the dandelions and buttercups
  reached across the green pastures.
Tee whee and tee whee came on the breezes, and the grackles
  chuzzled their syllables.
And it was the leaves with a strong soft wind over them
  that talked most of all and said more than any others
  though speaking the fewest words.
It was the green leaves trickling out the gaunt nowhere
  of winter, out on the gray hungry branches—
It was the leaves on the branches, beginning to be heavy,
  who said as they said one time before, “This is the be-
  ginning of summer.”

We shall never blame the birds who come
  where the river and the road make the Grand Crossing
  and talk there, sitting in circles talking bird talk.
If they ask in their circles as to who is here
  and as to who is not here and who used to be here,
Or if instead of counting up last year as against
  this year, they count up this year as against next
  year, and have their bird chatter about who is here
  this year who won’t be here next year,
We shall never blame the birds.

If I have put your face among leaf faces, child,
Or if I have put your voice among bird voices,
Blame me no more than the bluejays.

“Bird Talk” by Carl Sandburg—Public domain.

Question

Throughout the poem, the poet conveys the speaker’s point of view by

  • Agradually moving the speaker’s attention from one element of nature to another.
  • Bproviding the speaker’s motivation for describing the landscape in great detail.
  • Chighlighting the speaker’s emotional response as each season passes.
  • Dshifting the speaker’s focus from external observations to personal reflection.
Show the answer

Correct answer: D

This question asks for an explanation of how the poet conveys the speaker’s point of view on the natural world and the change of seasons as presented in the poem.

A. Incorrect. Though in the first stanza the speaker describes various elements of nature such as branches, frogs, birds, flowers, and leaves, these elemental shifts are not precisely gradual and occur only in the first stanza. Further, the first stanza, though appreciative of the natural world, is primarily descriptive and does not develop the speaker’s perspective to the degree that other parts of the poem do.

B. Incorrect. While it may be inferred that the speaker describes the landscape to such a degree because it has revealed a seasonal shift, that this is indeed the speaker’s motivation is never explicitly conveyed by the poet.

C. Incorrect. Multiple seasons do not pass in the poem; only the shift to the “ ‘beginning of summer’ ” (line 3) occurs in an endless cycle. Further, the speaker does not reveal an emotional response; it is the responses of nature that the speaker describes.

D. CORRECT. The first stanza is the speaker’s observations of the natural world embracing a new season. In the second stanza, the speaker begins to reflect on how the natural world, as embodied by the birds, calmly accepts seasonal change in its perpetual sameness. The reflection continues in the third stanza, where it is revealed that the speaker has been addressing these observations to a child.

Question analysis

This point-of-view question asks how the poet of “Bird Talk” conveys the speaker’s point of view throughout the poem. Students must keep in mind the long opening description of early summer and the two shorter stanzas after it, and decide which description of the poet’s method holds from the first line to the last.

NYS standards
  • 7R6 point of view and author position
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 point-of-view questions often ask what a speaker thinks. This one asks how the poet conveys that view across all three stanzas of a long free-verse poem, and the choices name methods in general terms, closer to grade 8 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from the whole text
Reading level
Passage: 296 words (a poem; reading level not measured); question and choices: about grade 10, 54 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The first stanza takes up 19 of the poem’s 32 lines and moves steadily from branches to frogs, birds, flowers and leaves, so a choice that describes it alone feels right. The point of view shows only when the poem turns to “We” and then to “I” addressing a child, and that third stanza is three hard lines. Students must keep the whole poem’s course in mind and choose the method that holds from first line to last.

  1. Evidence in several places moderate Working it out

    The answer depends on all three stanzas at once: the long catalog of the season in lines 1–19, the shift to “We shall never blame the birds,” and the closing “If I have put your face … child.” (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

  2. A very close wrong answer moderate Choosing the answer

    The choice about moving from one element of nature to another and the key both describe a shift in the speaker’s attention; they differ in what shifts, nature to nature or the outside world to the speaker’s own thoughts. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. How, not what moderate Working it out

    The stem asks how the poet conveys a point of view, a method, so a choice must describe what the speaker does across the poem, not what the speaker sees. (Anderson & Krathwohl, 2001)

  4. Figurative or literary language minor Understanding the question

    The last stanza is cryptic, “Blame me no more than the bluejays,” and students who miss that the speaker is now talking about putting a child into the poem lose the move to reflection. (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

Strategy: Write one phrase for what the speaker attends to in each stanza, then keep only a choice that describes the whole sequence, not one stanza.

Traps

  • Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks about point of view: how the speaker, the voice talking in the poem, sees things. It also asks how the poet shows that view across the whole poem. So look at what the speaker pays attention to in each of the three stanzas. A stanza is a group of lines set apart by a blank line.

  2. Follow the speaker through the stanzas

    • Stanza 1, lines 1–19: the speaker describes the world outside. There are branches, frogs, birds, flowers and leaves at the start of summer.
    • Stanza 2, lines 20–29: the speaker starts to judge and think. “We shall never blame the birds.”
    • Stanza 3, lines 30–32: the speaker says “I” and talks to a child: “If I have put your face among leaf faces, child.”

  3. Name the change in focus

    The poem starts with external observations, or things the speaker sees and hears outside. It ends with personal reflection, or the speaker’s own thoughts. In the last stanza, the speaker admits to putting a child’s face and voice into the poem. The speaker asks to be blamed “no more than the bluejays.” Since “We shall never blame the birds,” that means not at all. This move from looking outward to looking inward is how the speaker’s view comes through.

  4. Check each choice

    • A does not fit. Moving from branches to frogs to birds happens only in stanza 1. It misses the turn to “We” and “I.”
    • B does not fit. The speaker never says why the landscape is described.
    • C does not fit. Only one change of season happens, the start of summer. The anxious feelings belong to the birds, not the speaker.
    • D fits. The speaker moves from describing nature to thinking about it.

    Watch out: A describes stanza 1 very well. But the question says “Throughout the poem,” so the answer must cover all three stanzas.

  5. Choose the answer

    The speaker starts by watching nature and ends by thinking aloud to a child. So the answer is D, “shifting the speaker’s focus from external observations to personal reflection.”

Why this answer and not the closest wrong one

A is the closest wrong choice. Stanza 1 really does move from one part of nature to another. But that is only 19 of the 32 lines. In stanzas 2 and 3, the speaker turns to “We” and then “I,” and shares personal thoughts. Only D describes the speaker’s path from the first line to the last.

More questions like this

The original DOE question
Question 24 as shown on the DOE’s digital test, page 37 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C English, page 37, question 24.
The passage “Bird Talk” as printed (page 32)The passage “Bird Talk” as printed on page 32 of the DOE handbook.

Question 2 of 6

2024 Test B English, question 27
Passage Excerpt from Do Them No Harm!

Excerpt from Do Them No Harm!

by Zoa L. Swayne

This narrative is about the Nez Perce, an American Indian tribe, in what is now northern Idaho. The tribe is preparing for a gathering before the coming winter.

In the moon of Ta-Yum, the hottest days of summer, when salmon spawn in the little streams and huckleberries ripen in the high mountains, people from many villages of the Chopunnish Nation gathered in the Oyaip Prairie for the work and festivities of their annual camas1 harvest.

From far and near The People came. . . .

Should a stranger enter their homeland and ask, “Where are you from?” the reply was always, “We are Nee-mee-poo, The People who live here in this place.”

It was a time of Lawtiwa-mah-ton—a time of being friends together—when The People came for this last chance to enjoy being together before the Cold Moons kept them close to their fires. The visiting and trading, the foot racing and horse racing, and the gambling and stick games would be remembered and talked about long after they had forgotten the drudgery of digging and roasting camas, picking berries, or drying meat and fish. Lawtiwa-mah-ton! It was good to be friends together.

As was their custom since wahk-kee-ma, a time far back beyond the memory of man, they set up their camps in the same locations their parents and grandparents had occupied before them. Red Bear’s people, from Kamiah, made their camp near the trail that came out of the mountains. Their neighbors in Kamiah Valley, The People from Tee-e-lap-a-lo, had their camp close by. Across the wide meadow, by the great roasting pits, the camps of the Te-wap-poo and Ask-kah-poo were located. The tepees and ish-nash, brush shelters, of other groups nestled in their accustomed areas in and among the pines in such numbers that they encircled the entire meadow land.

Red Bear’s people had traveled all summer with neighboring bands, gathering and preparing roots, picking and drying berries, drying and smoking meat and fish for their winter food supply. Now they were at the Oyaip camp. The women worked hard to dig and cure as many bags of roots as they could during the warm, sunny days, for the sharp night air brought warnings that WARM was going and COLD was coming.

Everyone helped in some way. Most of the men fished or hunted for meat. While many of the women dug and roasted camas, other women and older children picked and dried berries.

And the younger children played. They played at hunting. They played with the babies. They played with their horses and puppies. They learned how to live through their play.

This sun, happiness, peace, and quiet blessed the Red Bear camp. All were busy with their daily tasks, until sudden cries came from the children playing by the trail.

“People coming! People coming! People coming on the trail from the high mountains!” they called as they ran to their elders, who looked sharply at the figures of approaching horsemen.

Were they friends or enemies? Did they bring good news or bad?

“Who can it be? What brings them here?” were the questions in every mind.

“Could they be the four hunters who had gone to Buffalo Country two summers past? Would they have news of the families who had gone long ago to Buffalo Country and never returned?”

It was customary for a hunting party to be gone for more than one season.

“Looks like hunters,” the older men agreed. “Looks like they had good hunting. Maybe our four hunters. [They have] been gone many moons.”

“Looks like five people—not four,” others observed.

Excitement grew as the riders came close enough to be recognized.

“A-a-a-a-a, they are our four hunters! But who is the fifth person?” they asked.

“Looks like a woman. Who is she?”

The hunters rode up to the welcoming crowd, proud to show off the loads of meat, hides, and other trophies of their hunt. They paraded around the encampment for all to see how strong their Hunting Power had been—what great hunters they, themselves, were.

Red Bear’s people rejoiced at their hunters’ success. Good hunters brought good to everybody. The meat meant plenty of food and the hides meant soft-tanned robes to give comfort through the Cold Moons. But it was the sight of the frail figure of the woman that aroused their curiosity. Who was she? Where had she come from? . . .

“Belongs to Red Bear people. Gone then come back,” the hunters said, as they dismounted and unloaded their packs. . . .

Now they could see! She was the daughter of the family gone so long ago! The girl-child who had left came back now—a grown woman.

“Wat-ku-ese!” the women cried. “Gone-from-Home-then-Come-Back. Wat-ku-ese!” And Wat-ku-ese was her name from that time on.

Gentle arms lifted Wat-ku-ese from her horse. The women brought her food and made a place for her to rest. For many suns they cared for her until she became stronger.

One evening Wat-ku-ese told her story for all to hear.

From DO THEM NO HARM!: Lewis and Clark Among the Nez Perce by Zoa L. Swayne. Published by Caxton Press. Copyright © 1990 by Zoa L. Swayne. Orofino, Idaho and Legacy House, Inc. Orofino, Idaho. All rights reserved.

1camas: a wild, edible plant

Question

How do paragraphs 11–13 affect the plot of the excerpt?

  • AThey establish a feeling of mystery by presenting different points of view about the arriving people.
  • BThey contribute to the rising action by developing the idea that the arriving people could present a problem.
  • CThey lead to a turning point by describing the moment that the crowd realizes why the strangers have come.
  • DThey explain the cause of the main conflict by providing background information about the strangers.
Show the answer

Correct answer: B

The question asks how paragraphs 11–13 affect the plot of the excerpt.

A. Incorrect. The paragraphs contain questions about the arriving people. They do not present differing points of view.

B. CORRECT. The paragraphs ask questions in which the characters wonder whether the arriving people are friends or enemies, or whether they might be bringing good news or bad news. By suggesting that the arriving people could present problems such as these, the paragraphs contribute to the rising action.

C. Incorrect. The paragraphs do not lead to a turning point. Instead, they lead to the climax—the moment the characters discover who the visitors are. This means the paragraphs contribute to the rising action.

D. Incorrect. Paragraphs 11–13 ask questions. They do not provide background information about the strangers; instead, they ask for it.

Question analysis

This plot question asks how paragraphs 11–13, a series of questions the villagers ask as riders approach their camp, affect the plot of the Nez Perce narrative. Students must know terms such as rising action and turning point and apply them precisely, since several choices describe these paragraphs in ways that sound right but place them wrongly in the story.

NYS standards
  • 7R3 plot elements and ideas
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 plot questions usually ask what one event causes. Here students apply terms such as rising action and turning point to three short paragraphs of questions, which matches the grade 8 focus on how parts of a story propel the action.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: about grade 5, 885 words; question and choices: about grade 9, 77 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Paragraphs 11–13 contain nothing but questions, so the answer depends on what questions do in a plot. Friends or enemies, good news or bad, they raise the chance of trouble and build toward the arrival. The trap is that they also create mystery, and the choice that says so fails only on one phrase about differing viewpoints. A second risk is jumping ahead: at this point no one even knows who the riders are, let alone why they came.

  1. A very close wrong answer major Choosing the answer

    The right choice and the one about mystery both see the questions building uncertainty about the riders; only the claim of “different points of view,” when the questions are “in every mind,” separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  2. How, not what moderate Working it out

    The stem asks what paragraphs 11–13 do for the plot, so students must place a run of unanswered questions in the story’s shape rather than retell them. (Anderson & Krathwohl, 2001)

  3. Something you must already know minor Working it out

    The choices rely on plot terms the excerpt never explains, rising action, turning point and main conflict, and judging them means knowing what each one names. (Anderson & Krathwohl, 2001)

Strategy: Note what the villagers know and do not yet know in paragraphs 11–13, then compare that with what happens later in the excerpt to place these paragraphs in the plot.

Traps

  • Half right: One half of a two-part choice is right and the other is not.
  • Wrong part of the text: A choice is true of another paragraph, stanza or speaker than the one asked about.
  • Wrong purpose or structure: Names a purpose, structure or effect that the part does not serve, or describes what a part says when the question asks what it does.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    This asks how paragraphs 11–13 move the story along. You need some plot words. The plot is the order of events in a story. The rising action is the part where tension builds. Tension is the feeling that something big may happen. A turning point is the moment when things change.

  2. Read paragraphs 11 to 13

    These paragraphs are all questions. Paragraph 11 asks: “Were they friends or enemies? Did they bring good news or bad?” In paragraph 12, “Who can it be?” is one of “the questions in every mind.” Paragraph 13 asks if the riders are the hunters who left “two summers past.”

  3. Decide what the questions do

    The words “enemies” and “bad” news show the camp is worried. The riders might bring trouble. No one knows yet. So the questions build tension as the riders get closer. That is rising action. The answers come later. The crowd knows the hunters in paragraph 18. They learn who the woman is in paragraphs 22 and 23.

  4. Check each choice

    • B fits. The questions raise the chance of a problem and build tension.
    • A: there is mystery, but not “different points of view.” Everyone asks the same questions.
    • C: in these paragraphs, no one knows who the riders are yet, let alone why they came. So nothing here is the moment the crowd realizes anything.
    • D: the paragraphs ask about the strangers. They give no background facts about them.

  5. Pick the answer

    Watch out: A is very close. The questions do create mystery. But one wrong phrase, “different points of view,” rules it out. So the answer is B.

Why this answer and not the closest wrong one

A is the closest wrong answer, because the questions do make the scene mysterious. But the questions are “in every mind,” so there are no different points of view. B names what the questions do for the plot: they hint that the riders could bring trouble, and they build tension.

More questions like this

The original DOE question
Question 27 as printed on page 180 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 180, question 27. Open page 180 of the handbook (PDF)
The passage “Excerpt from Do Them No Harm!” as printed (pages 177–178)The passage “Excerpt from Do Them No Harm!” as printed on page 177 of the DOE handbook.The passage “Excerpt from Do Them No Harm!” as printed on page 178 of the DOE handbook.

Question 3 of 6

2026 Practice Test C English, question 39
Passage Excerpt from “Growing Peppers on the ISS Is Just the Start of Space Farming”

Excerpt from “Growing Peppers on the ISS Is Just the Start of Space Farming”

by Melanie Canales

The author, Melanie Canales, is a first-generation Peruvian American farmer and beekeeper based in Virginia’s Shenandoah Valley, on lands of the indigenous Monacan and Manahoac tribes. Her work focuses on creating healthy relationships between people and their surrounding ecosystems.

Unburdened by the constraints of gravity, red and green peppers jut out at 45-degree angles inside the Advanced Plant Habitat (APH), a sort of space terrarium not much larger than a microwave. Four chile pepper plants stand effortlessly upright, despite the dozens of glossy fruits weighing them down. These plants have lived entirely in space; their leaves have never been chewed on by insects or rustled by a summer breeze, [and] their stems are unfamiliar with bending toward the sun’s arc across the sky. Scissors glint under the tank’s white and blue lights as astronaut Mark Vande Hei and his team snip the stems of those that are ready for harvest. The peppers whirl around their heads until the astronauts catch them and tape them against a board to photograph.

Back on Earth, the Plant Habitat-04 team of engineers and plant scientists are observing and conferring with the astronauts. Of the 26 peppers in this batch, only the 14 finest will stay on the International Space Station [ISS] for consumption. The rest will be wrapped in foil, sealed in a Ziploc bag, then frozen at a brisk −80 degrees, until they can come roaring back to Earth in the next cargo capsule to be studied later. Now, after a 137-day growth cycle, the astronauts remove the plants from the module and trash them. Project Plant Habitat-04 is complete. It’s taco night on the ISS.

Since 2014, NASA1 has experimented with growing lettuces, brassicas,2 and zinnias3 in space, an endeavor that relies on highly specialized technology over 50 years in the making. [Fall 2021’s] two successful pepper harvests . . . will provide data on the nutritional and psychological benefits of growing vegetables on-craft, as well as a crop’s ability to reliably produce long-term in microgravity. While controlled environmental agriculture is not new, the APH experiment represents an evolution in specialized growth habitats. It doesn’t aim to re-create Earth’s conditions, but to perfect each isolated variable of plant growth in the clinical environment of a spaceship.

“The Advanced Plant Habitat is the most complex plant growth system on orbit today,” says Lashelle Spencer, a plant scientist at NASA’s Kennedy Space Center. Its more than 180 sensors control and monitor temperature, humidity, and carbon dioxide. The astronauts can adjust the color and intensity of the light, and how much moisture the plants’ roots are getting. It waters itself.

It’s the day after Thanksgiving, and Spencer has been at Kennedy since 5 a.m. to facilitate the peppers’ final harvest. As part of the project team, she played a crucial role in preparing the seeds that were sent hurtling into space in June and guiding the astronauts through maintaining the plants in orbit. . . . Though astronauts can spend upwards of 100 days in space, their on-mission meals come dehydrated and pre-packaged; their vitamins and minerals are isolated in supplements, which lose nutritional value the longer they’re stored. Spencer’s goal is to create the conditions necessary for cultivating healthy plants in space, so those plants can sustain healthy astronauts on long-term missions. Astronaut food is great, she says—“especially the shrimp cocktail. But you’re missing that crunch. You’re missing that fresh pop of flavor, the green flavor that’s not there in that packaged food.”

The sensory experience of growing productive crops can also help mitigate4 the psychological effects of long-term space travel. . . . Spencer says the team cracked open the door of the APH every day to observe their vegetable companions with all the tenderness of home gardeners. When harvest day came, they batted their bounty around the ISS, taking selfies and delighting in watching the fruits pirouetting around the spacecraft. . . .

“We were thinking no heat, so that [the peppers] wouldn’t be dangerous, but maybe the astronauts need a little spice in their life,” says Paul Bosland, who along with his colleagues at the Chile Pepper Institute genetically engineered5 the NuMex Española Improved chile pepper seeds grown in Plant Habitat-04. . . .

Working with NASA, Bosland cultivated a variety that could accommodate both the nutritional needs of astronauts as well as the logistics of growing a plant in space. . . .

. . . The experiment is helping gather the data necessary to determine the nutritional content of crops grown in space, and therefore how many people they can feed, and for how long. Though much is still unknown, Spencer is certain of one thing that future astronauts will need to do: “I think in an optimal world, a scientist like me would say that they would be growing plants from day one. From the day they left to the day they came back, they would be growing them.”

From “Growing Peppers on the ISS Is Just the Start of Space Farming” by Melanie Canales from WIRED, December 21, 2021. Copyright © 2021 Condé Nast. All rights reserved.

1NASA: National Aeronautics and Space Administration

2brassicas: a large genus of herbs, such as broccoli and cabbage

3zinnias: a genus of herbs and low shrubs that have flowers at their heads

4mitigate: to make something less severe or harmful

5genetically engineered: combined parts of DNA from different plants to produce a new plant

Question

The description in paragraph 5 of Lashelle Spencer’s goal develops a central idea of the excerpt by

  • Aillustrating that certain types of packaged foods available in space lack particular flavors.
  • Bsupporting the notion that spending more than 100 days in space is detrimental to astronauts’ health.
  • Coffering an example of how the limited selection of on-mission meals becomes less appealing over time.
  • Dmaking an important point about the astronauts’ current reliance on supplements for vitamins and minerals.
Show the answer

Correct answer: D

The question asks how the description of Lashelle Spencer’s goal in paragraph 5 develops a central idea of the excerpt.

A. Incorrect. Although Spencer mentions how astronauts miss certain eating experiences like the “fresh pop” and “green flavor” not found in packaged and dehydrated food, a central idea of the excerpt is that there is a need to provide fresh, healthy foods for long missions, not that food flavor should be improved.

B. Incorrect. While the excerpt focuses on how spending more than 100 days in space affects nutrition, it does not discuss specific ways in which spending long periods of time in space is detrimental to the health of astronauts.

C. Incorrect. Although paragraph 5 states that astronauts “can spend upwards of 100 days in space,” the excerpt does not develop a central idea that existing meals in space become less appealing over time.

D. CORRECT. The paragraph explains that astronauts currently rely on supplements for vitamins and minerals, which lose their nutritional value over time. Spencer’s goal for growing plants in space is to “sustain healthy astronauts,” supporting a central idea that astronauts need more nutritionally valuable food to sustain them while in space for long periods of time.

Question analysis

Lashelle Spencer’s goal for growing plants in space is described in paragraph 5 of the WIRED excerpt, and this central-idea question asks how that description develops a central idea of the whole excerpt. Students must connect a single sentence about a scientist’s aim to the article’s larger point, and every choice draws on details from that same paragraph.

NYS standards
  • 7R2 theme or central idea
  • 7R5 structure and genre features
Grade level
Grade 8 reading and task
Compared with grade 7
Above grade 7 Grade 7 students trace how a central idea develops, usually through a plainly stated example. Here the goal’s connection to the central idea runs through the sentences around it, and the choices are abstract restatements, which asks more than typical grade 7 work.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Evidence
Answered from several paragraphs or stanzas
Reading level
Passage: about grade 10, 890 words; question and choices: about grade 12, 75 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every choice draws on paragraph 5, and none mentions growing plants, which leaves students looking for the goal without finding it. The flavor quotation that follows the goal is the most memorable detail and the easiest wrong pick. The key works only through context: the goal answers the problem the previous sentence sets out, crews depending on supplements that weaken in storage, which is why it serves the article’s case for fresh food on long missions.

  1. Reading between the lines moderate Working it out

    Spencer’s goal names healthy plants and healthy astronauts but no supplements; its link to the key runs through the sentence before it, where vitamins sit in supplements that “lose nutritional value the longer they’re stored.” (Graesser, Singer & Trabasso, 1994; Kintsch, 1998)

  2. Every choice must be checked moderate Choosing the answer

    No choice restates the goal of growing plants itself, so none can be matched at a glance; each must be tested for whether it ties the goal to the excerpt’s concern with nutrition on long missions. (Embretson & Wetzel, 1987)

  3. How, not what minor Working it out

    The question asks how the description develops a central idea, its role in the article, so knowing what Spencer wants is only the first step toward the answer. (Anderson & Krathwohl, 2001)

  4. Evidence in several places minor Working it out

    The central idea must come from the whole excerpt: paragraph 3’s “nutritional and psychological benefits,” paragraph 8’s “nutritional needs of astronauts” and paragraph 9’s “nutritional content of crops.” (Kirsch & Mosenthal, 1990; Gorin & Embretson, 2006)

Strategy: State a central idea of the whole excerpt in your own words, reread Spencer’s goal with the sentences around it in paragraph 5, and pick the option that connects the goal to that central idea.

Traps

  • Too narrow: A single detail offered as the central idea or a summary, or a summary that keeps minor details and leaves out an essential part.
  • Twists what the text says: Uses the text's topic and ideas but changes what it says: who did it, how much, when, or why.
  • Reads too much in: A reasonable-sounding conclusion that goes further than the text allows.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Working it out, step by step

  1. Figure out what the question asks

    A central idea is a main point the whole article keeps coming back to. This question asks how Spencer’s goal, in paragraph 5, helps build one of those main points. So first, find the goal. Then connect it to the big idea.

  2. Find the goal and the sentence before

    Paragraph 5 says: “Spencer’s goal is to create the conditions necessary for cultivating healthy plants in space, so those plants can sustain healthy astronauts on long-term missions.” Cultivating means growing. The sentence just before it shows why she has this goal. Astronauts’ “vitamins and minerals are isolated in supplements, which lose nutritional value the longer they’re stored.”

  3. Connect the goal to the big idea

    Today astronauts get vitamins from pills, called supplements. Those pills lose their health value as they sit. Spencer wants fresh plants grown in space to keep astronauts healthy instead. So her goal points to a problem: relying on weak pills. The article returns to food and health again and again. Paragraph 8 mentions “the nutritional needs of astronauts.” Nutritional means having to do with healthy food.

  4. Check each choice

    • A is wrong. The “fresh pop of flavor” line comes after the goal. It is about taste, a side point.
    • B is wrong. The article never says long trips in space harm astronauts’ health.
    • C is wrong. It is the stored supplements that lose value over time, not meals that get less tasty.
    • D is right. The goal answers a problem: astronauts depend on supplements that weaken in storage. That ties to the article’s idea that astronauts need healthy fresh food on long trips.

    Watch out: Spencer’s flavor quote is the most memorable line nearby, so A pulls hard. But her goal is about health, not taste.

  5. Choose the answer

    Spencer wants healthy plants because pills are not enough. So the answer is D, “making an important point about the astronauts’ current reliance on supplements for vitamins and minerals.” Reliance means depending on something.

Why this answer and not the closest wrong one

A is the closest wrong answer. Spencer really does miss the crunch and flavor of fresh food. But her goal is to “sustain healthy astronauts,” which is about health. Health is the big idea the article keeps returning to. D connects the goal to that idea through the supplements that lose value.

More questions like this

The original DOE question
Question 39 as shown on the DOE’s digital test, page 57 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C English, page 57, question 39.
The passage “Excerpt from “Growing Peppers on the ISS Is Just the Start of Space Farming”” as printed (pages 50–51)The passage “Excerpt from “Growing Peppers on the ISS Is Just the Start of Space Farming”” as printed on page 50 of the DOE handbook.The passage “Excerpt from “Growing Peppers on the ISS Is Just the Start of Space Farming”” as printed on page 51 of the DOE handbook.

Question 4 of 6

2026 Practice Test C Math, question 35

Question

If 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a, what is the value of a in terms of b?

Enter your answer in the space provided. Enter only your answer.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: a = 10b

To find the value of a in terms of b, solve 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a for a. Begin by distributing 4 on the left side of the equation:

(4×a)+(4×2b)=28b+2a(4 \times a) + (4 \times 2b) = 28b + 2a

4a+8b=28b+2a4a + 8b = 28b + 2a

Then, get the a terms on one side and the b terms on the other:

4a−2a=28b−8b4a - 2a = 28b - 8b

Solve for the value of a in terms of b by combining like terms and then dividing each side by 2:

2a=20b2a = 20b

a=10ba = 10b

Question analysis

An algebra question gives an equation with a and b on both sides, one side written as 4 times a sum in parentheses, and students type a in terms of b after a printed "a =". It tests the distributive property and solving linear equations. The answer is typed, so a slip in distributing or moving terms is not caught by any choices.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.EE.1 linear expressions with rational coefficients
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 students solve equations like 4(x + 2) = 28 with numbers. This one has letters on both sides, needs the distributive property and like terms, and asks for a in terms of b, which is grade 8 algebra.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Solution
3 steps
Reading level
Question and choices: about grade 4, 23 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Every step is routine, yet there are four of them, each with a classic slip, and the typed answer catches none. Multiplying the 4 into only the a, or carrying 8b across without changing its sign, gives another whole multiple of b that looks finished. Testing b = 1, where 4(a + 2) and 28 + 2a must come out equal, confirms an entry quickly.

  1. Variables instead of numbers moderate Understanding the question

    Both letters appear on both sides, and the answer must stay a multiple of b, so there is no number to check an entry against unless students invent one. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)

  2. Many steps moderate Working it out

    Four moves run in a chain, distributing the 4, gathering the a terms, gathering the b terms and dividing by 2, and an error at any one carries into the typed result. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. An answer in an unexpected form minor Choosing the answer

    The answer is typed after a printed “a =”, with no choices, so 20b, the line before dividing by 2, can be entered without anything on screen questioning it. (Enright, Morley & Sheehan, 2002)

Strategy: Distribute the 4 to both terms in the parentheses, gather the a terms on one side and the b terms on the other, then divide by the coefficient of a.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
k(x+y)=kx+kyk(x + y) = kx + ky
k is the number outside the parentheses. x and y are the terms inside. The number outside multiplies every term inside, not just the first one. It works because 4 groups of (x + y) hold 4 x’s and 4 y’s.
Doing the same thing to both sides
if A=B, then A−C=B−C and A2=B2\text{if } A = B, \text{ then } A - C = B - C \text{ and } \frac{A}{2} = \frac{B}{2}
A and B are the two equal sides of an equation. C is any amount. Taking the same amount from both sides, or dividing both by the same number, keeps them equal, like a balanced scale.

  1. Sort out the question

    The equation is 4(a+2b)=28b+2a4(a + 2b) = 28b + 2a. You need a in terms of b. That means a alone on one side, with only b and numbers on the other.

    The answer box already shows a =, so you type only what comes after it.

  2. Multiply out the parentheses

    Use the distributive property. The 4 multiplies both terms inside:

    4×a+4×2b=28b+2a4 \times a + 4 \times 2b = 28b + 2a

    4a+8b=28b+2a4a + 8b = 28b + 2a

    Watch out: multiplying only the a by 4 gives 4a + 2b. That leads to the wrong answer 13b.

  3. Gather the a terms

    Like terms are terms with the same letter, such as 4a and 2a. Put the a terms on the left. Subtract 2a from both sides:

    4a−2a+8b=28b4a - 2a + 8b = 28b

    2a+8b=28b2a + 8b = 28b

  4. Gather the b terms

    Now subtract 8b from both sides:

    2a=28b−8b2a = 28b - 8b

    2a=20b2a = 20b

    Watch out: moving 8b without changing its sign gives 28b+8b=36b28b + 8b = 36b. That leads to 18b, which is wrong.

  5. Divide both sides by 2

    2a means 2 times a. Divide both sides by 2 to get a alone:

    a=20b2=10ba = \frac{20b}{2} = 10b

    Watch out: 20b is equal to 2a, not a. Do not stop there.

  6. Enter it after the printed a =

    On the test, the box already shows a =. Press 1, then 0, then the b key. You do not need a times sign. Do not type a = again. On this site, type 10b in the box.

    So you would enter 10b, which makes a = 10b.

Check your answer

Pick b = 1, so a = 10. The left side is 4(10+2)=4×12=484(10 + 2) = 4 \times 12 = 48. The right side is 28+2×10=4828 + 2 \times 10 = 48. The sides match, so 10b works.

More questions like this

The original DOE question
Question 35 as shown on the DOE’s digital test, page 108 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test C Math, page 108, question 35.

Question 5 of 6

2024 Test B Math, question 14

Question

Karen reads at the rate of 60 pages per day. Martina reads 25% more pages per day than Karen. If the two students begin 1,500-page reading assignments on the same day, how many days sooner will Martina complete the assignment than Karen?

  • A4
  • B5
  • C7
  • D15
Show the answer

Correct answer: B

First, determine the rate at which Martina reads, in pages per day, using the rate at which Karen reads, 60 pages per day.

60+60∙25100=60+60\bullet\frac{25}{100}=

60+60∙0.25=60+60\bullet 0.25=

60+15=7560+15=75

Next, determine the number of days it will take each person to complete the 1,500-page reading assignment.

Karen: 1,500 pagesk days=60 pages1 day\frac{1{,}500\text{ pages}}{k\text{ days}}=\frac{60\text{ pages}}{1\text{ day}}

60k=1,50060k=1{,}500

(60÷60)k=1,500÷60(60\div 60)k=1{,}500\div 60

k=25 daysk=25\text{ days}

Martina: 1,500 pagesm days=75 pages1 day\frac{1{,}500\text{ pages}}{m\text{ days}}=\frac{75\text{ pages}}{1\text{ day}}

75m=1,50075m=1{,}500

(75÷75)m=1,500÷75(75\div 75)m=1{,}500\div 75

m=20 daysm=20\text{ days}

Finally, determine the difference between the two numbers of days.

25−20=525-20=5

So Martina will complete the assignment 5 days sooner than Karen.

Question analysis

A reading-rate word problem: Karen reads 60 pages a day, and Martina reads 25% more pages a day. Both start a 1,500-page assignment on the same day, and students find how many days sooner Martina finishes. It combines a percent increase with rates and times, and asks for a difference in days.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.RP.3b unit rate problems
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually stop after one increase or decrease. This one turns the percent into a new rate, finds two finishing times and compares them, four steps that go beyond a typical grade 7 task.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
4 steps
Reading level
Question and choices: about grade 6, 40 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is easy on its own; the difficulty is in carrying the chain to the end. Three of the choices are numbers a student meets or makes along the way: the extra pages per day, a quotient of two rates, or a rate built on 25 pages instead of 25%. Writing both finishing times down before subtracting keeps the target, a number of days, in view.

  1. Many steps moderate Working it out

    Four operations chain together: 25% of 60, Martina's 75 pages, the two finishing times 1,500 ÷ 60 and 1,500 ÷ 75, and finally their difference. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. Combines ideas moderate Working it out

    A percent increase in a daily rate has to be turned into a change in days, and 25% more pages a day does not mean 25% fewer days. (Embretson, 1998; Daniel & Embretson, 2010)

  3. Turning words into math minor Understanding the question

    “25% more pages per day than Karen” makes Martina's rate 1.25 × 60, and “how many days sooner” asks for a difference of two times, not of two rates. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Increase 60 by 25% to get Martina's daily pages, divide 1,500 by each girl's daily rate to get her number of days, then subtract the two numbers of days.

Traps

  • Stops one step early: A choice is a number from partway through the solution.
  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Percent increase
new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}
Old is the starting amount, and p is the percent it grows by. First find p percent of the old amount. That is the extra. Then add the extra to the old amount to get the new amount.
Time from a rate
days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}
Total pages is the whole assignment. Pages per day is how fast someone reads. Dividing tells how many days of reading fit into the whole job.

  1. Sort out what you know

    Karen reads 60 pages per day. Martina reads 25% more pages per day than Karen. Each girl must read 1,500 pages. You need how many days sooner Martina finishes. That is the difference between their numbers of days.

  2. Find Martina’s pages per day

    25% more means Karen’s 60 pages plus 25% of 60.

    new=old+p100×old\text{new} = \text{old} + \frac{p}{100} \times \text{old}

    60+25100×60=60+15=7560 + \frac{25}{100} \times 60 = 60 + 15 = 75

    Martina reads 75 pages per day. Watch out: 15 is extra pages, not days. Stopping here gives 15, choice D. Adding 25 pages instead of 25% gives 85 pages and leads to 7, choice C.

  3. Find each girl’s number of days

    Divide the whole assignment by each girl’s pages per day.

    days=total pages÷pages per day\text{days} = \text{total pages} \div \text{pages per day}

    Karen: 1,500÷60=25\text{Karen: } 1{,}500 \div 60 = 25

    Martina: 1,500÷75=20\text{Martina: } 1{,}500 \div 75 = 20

    Karen needs 25 days, and Martina needs 20 days.

  4. Subtract the days

    “How many days sooner” asks for the difference.

    25−20=525 - 20 = 5

    Martina finishes 5 days sooner. Watch out: dividing 60 by 15 mixes two rates and gives 4, choice A.

    So the answer is B, 5.

Check your answer

Multiply back. Karen: 25×60=1,50025 \times 60 = 1{,}500 pages. Martina: 20×75=1,50020 \times 75 = 1{,}500 pages. Both read all 1,500 pages, and 25 days is 5 days more than 20 days.

More questions like this

The original DOE question
Question 71 as printed on page 202 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 202, question 71. Open page 202 of the handbook (PDF)

Question 6 of 6

2024 Test B Math, question 56

Question

A rectangular field had a length of 80 meters and a width of 50 meters. The field was recently enlarged, with the length increased by 15% and the width increased by 10 meters. By what percentage was the area of the field increased?

  • A26.5%
  • B35%
  • C38%
  • D42.5%
Show the answer

Correct answer: C

The original area of the field was 4,000 square meters.

80×50=4,00080\times 50=4{,}000

When the field was enlarged, the length increased by 15%, so the new length is 92 meters.

80×15%80\times 15\%

80×0.15=1280\times 0.15=12

80+12=9280+12=92

The width increased by 10 meters, so the new width is 60 meters.

50+10=6050+10=60

The new area of the field is 5,520 square meters.

92×60=5,52092\times 60=5{,}520

The percent increase is found by finding the difference between the two areas and dividing the increase by the original area.

5,520−4,000=1,5205{,}520-4{,}000=1{,}520

1,520÷4,000=0.38=38%1{,}520\div 4{,}000=0.38=38\%

Question analysis

A rectangular field 80 meters long and 50 meters wide has its length increased by 15 percent and its width by 10 meters. Students find the percent by which the field's area grew. It tests percent increase with rectangle area, and the tricky part is that one side changes by a percent and the other by a fixed length.

NYS standards
  • 7.RP.3 multistep ratio and percent problems
  • 6.G.1 area of polygons
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Above grade 7 Grade 7 percent problems usually apply one percent change to one quantity. This changes a length by a percent and a width by a fixed amount, then asks for the percent change in area, a five-step chain with two kinds of change.
Estimated difficulty
Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
What makes it hard
Solution
5 steps
Reading level
Question and choices: about grade 6, 39 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each step is ordinary, but the two changes are stated in different units, one a percent and one in meters, which tempts students to force them into the same form. Adding the two percents is the most natural shortcut and lands close to the right value, because it misses only the small extra area where both changes overlap. Computing the old and new areas directly is slower but leaves nothing to chance.

  1. Combines ideas moderate Working it out

    Percent change has to be combined with area: the length grows by a percent and the width by meters, and the question asks for the percent change of their product. (Embretson, 1998; Daniel & Embretson, 2010)

  2. Many steps moderate Working it out

    Five computations are chained: the new length 92, the new width 60, the areas 4,000 and 5,520, and the increase 1,520 as a share of 4,000. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  3. A very close wrong answer minor Choosing the answer

    The 35% from adding 15% and 20% sits only 3 points below 38%, so a rough estimate of the new area cannot rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Compute the new length and the new width, find the old and new areas, and divide the increase in area by the original area.

Traps

  • Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
  • Units or scale: The answer is wanted in other units, or a rate or scale must be converted first.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Area of a rectangle
A=l×wA = l \times w
A is the area, the flat space inside. l is the length and w is the width. Multiplying counts how many unit squares fit, row by row.
Percent increase
percent increase=new−oldold×100\text{percent increase} = \frac{\text{new} - \text{old}}{\text{old}} \times 100
old is the starting amount, and new is the amount after. new − old is how much it grew. Dividing by old compares the growth with the start. Multiplying by 100 turns that into a percent.

  1. Sort out the question

    The field was 80 meters long and 50 meters wide. The length grew by 15%, a percent. The width grew by 10 meters, a number of meters. These are two different kinds of increase.

    You need the percent by which the area grew.

  2. Find the old area

    You need the old area to compare with the new one.

    A=80×50=4,000A = 80 \times 50 = 4{,}000

    The old field covered 4,000 square meters.

  3. Find the new length and width

    15% of 80 is 0.15×80=120.15 \times 80 = 12. So the new length is 80+12=9280 + 12 = 92 meters.

    The new width is 50+10=6050 + 10 = 60 meters.

    Watch out: 10 meters is not 10%. Treating it as 10% leads to 26.5%, choice A. Treating 15% as 15 meters leads to 42.5%, choice D.

  4. Find the new area

    Multiply the new length by the new width.

    A=92×60=5,520A = 92 \times 60 = 5{,}520

    The new field covers 5,520 square meters. That is 5,520−4,000=1,5205{,}520 - 4{,}000 = 1{,}520 square meters more.

  5. Find the percent increase

    percent increase=5,520−4,0004,000×100=1,5204,000×100=38\text{percent increase} = \frac{5{,}520 - 4{,}000}{4{,}000} \times 100 = \frac{1{,}520}{4{,}000} \times 100 = 38

    The area grew by 38%. Watch out: The width grew by 20%, since 10 is 20% of 50. Adding 15% + 20% = 35% gives choice B. That misses the extra corner piece where both new strips meet.

    So the answer is C, 38%.

Check your answer

Use growth factors. The length is multiplied by 1.15. The width goes from 50 to 60, so it is multiplied by 1.2. The area is multiplied by 1.15×1.2=1.381.15 \times 1.2 = 1.38, which is 38% more.

More questions like this

The original DOE question
Question 113 as printed on page 215 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 215, question 113. Open page 215 of the handbook (PDF)

Sample #164360